chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions47. X ~ Bin(6, .10)a. P(X = 1) =⎛n⎞⎜ ⎟⎝ x⎠x n−x⎛6⎞1 5( p)(1 − p)= ⎜ ⎟(.1)(.9) = . 3543b. P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)].⎝1⎠66⎛ ⎞⎝0⎠0From a , we know P(X = 1) = .3543, and P(X = 0) = ⎜ ⎟(.1)(.9) = . 5314 .Hence P(X ≥ 2) = 1 – [.3543 + .5314] = .1143c. Either 4 or 5 goblets must be selected44⎛ ⎞⎝0⎠0i) Select 4 goblets with zero defects: P(X = 0) = ⎜ ⎟(.1)(.9) = . 6561.ii) Select 4 goblets, one of which has a defect, and the 5 th is good:⎡⎛4⎞1 3⎤⎢⎜⎟(.1)(.9) ⎥ × .9 = .26244⎣⎝1⎠⎦So the desired probability is .6561 + .26244 = .9185448. Let S = comes to a complete stop, so p = .25 , n = 20a. P(X ≤ 6) = B(6;20,.25) = .786b. P(X = 6) = b(6;20,.20) = B(6;20,.25) - B(5;20,.25) = .786 - .617 = .169c. P(X ≥ 6) = 1 – P(X ≤ 5) = 1 - B(5;20,.25) = 1 - .617 = .383d. E(X) = (20)(.25) = 5. We expect 5 of the next 20 to stop.49. Let S = has at least one citation. Then p = .4, n = 15a. If at least 10 have no citations (Failure), then at most 5 have had at least one (Success):P(X ≤ 5) = B(5;15,.40) = .403b. P(X ≤ 7) = B(7;15,.40) = .787c. P( 5 ≤ X ≤ 10) = P(X ≤ 10) – P(X ≤ 4) = .991 - .217 = .774108